y(f(x))=y(x)+x

[Maxim]

Is it possible to solve equation y(f(x))=y(x)+x if f(.) is a known
function? (e.g., f(x)=exp(x)-1 ?)

[Marc Nardmann]

Maxim wrote:

>Is it possible to solve equation y(f(x))=y(x)+x if f(.) is a known
>function? (e.g., f(x)=exp(x)-1 ?)
>
>

If f(0)=0 and f is differentiable at 0 with f'(0)=1, as in the case
f(x)=exp(x)-1, then there is no solution y which is differentiable at 0.

PROOF: Assume that y is differentiable at 0. Then so is y°f. For x not
equal to 0, we have [(y°f)(x)-(y°f)(0)]/x - [y(x)-y(0)]/x = 1; so for x
-> 0, we obtain 1+y'(0) = (y°f)'(0) = y'(f(0)) f'(0) = y'(0), a
contradiction.

-- Marc Nardmann

[Dave Rusin]

In article <vz6xjmdri49p@legacy>, Maxim <MOsa...@nes.ru> wrote:
>Is it possible to solve equation y(f(x))=y(x)+x if f(.) is a known
>function? (e.g., f(x)=exp(x)-1 ?)

What do you mean by "solve"?

Exactly this example is treated as a prototype for a broad family of
similar problems in a paper by Szekeres; he argues that there is one
particularly optimal solution y and in an accompanying paper provides
tables of numerical values. (He uses the analysis to describe a
one-parameter family of functions f_s with f_s o f_t = f_{s+t} and
in particular describes a functional "square root" f_{1/2} of f = f_1.)

MR0141905 (25 #5302)
Szekeres, G.
Fractional iteration of exponentially growing functions.
J. Austral. Math. Soc. 2 1961/1962 301--320. MSC section 39.99

dave

[Daniel Geisler]

See
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A052122
for f(x) such that f(f(x)) = exp(x)-1.
Daniel

[Alain verghote]

Dear All,

We may show that if it exists m |f(x)= m^[x](x) xth iterate of m(x)
equation y(f(x))=y(x)+x is equivalent to y(m(x)= y(x) + 1 (sci.math 09
june ).
So with f(x)=exp(x)-1 you have to find m^[x](x)= exp(x)-1 ,an harder
task than solving or computing m(x) from m(m(x))= exp(x)-1 ,

Sincerely,Alain.

[Alain Verghote]

Dear friends,

Very often we may build y(x)as a function point to point .
Here y(f(x))=y(x)+x then from y(x0) , y(f(x0))=y(x0)+x0 ....
in fact we've got:y(f^[n](x0)=y(f^[n-1](x0))+f^[n-1](x0) or
y(f^[n](x0)=y(x0)+x0 +f(x0)+f^[2](x0)+....
f^[n](x0) stands for f(f(f(f(f(f (x0)..))) f n times.
In our case we must choose x0 # 0 (to avoid y(0)=y(0)+0 a loop!).
two consequences:
value y(x0) is 'free',
we obtain a'gruyère',that is to say a holy swiss cheese:function is
known only in points x0, f(x0),....f^[n](x0).

Sincerely,Alain.